1955 AHSME Problems/Problem 17: Difference between revisions

Revision as of 16:12, 29 November 2020

If $\log x-5 \log 3=-2$ , then $x$ equals:

$\textbf{(A)}\ 1.25\qquad\textbf{(B)}\ 0.81\qquad\textbf{(C)}\ 2.43\qquad\textbf{(D)}\ 0.8\qquad\textbf{(E)}\ \text{either 0.8 or 1.25}$

Solution

Definitions and Properties

$\log n$ is defined as the value of $x$ that satisfies the equation $n = 10^x$ . Note that other bases can be applied as well, so $\log_b n$ would be defined as the answer to $n = b^x$

$y \log n = \log n^y$

$\log x - \log y = \log (x / y)$

Learn more properties here: [[1]]

Solving the Equation

$\log x-5 \log 3=-2$

$\log x- \log 3^5=-2$

$\log x- \log 243 =-2$

$\log x / 243 = -2$

$x/243 = 10^{-2}$

$x=\frac{243}{100}$ $x=\boxed{\textbf{(C) }2.43}$

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1955 AHSME Problems/Problem 17: Difference between revisions

Revision as of 16:12, 29 November 2020

Contents

Solution

Definitions and Properties

Solving the Equation

See Also